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For example

In a certain home there are A, B and C keyrings. The first with five keys, the second with seven and the third with eight, of which only one of each keyring opens the garage door. A keyring is chosen at random, and from it, a key to open the garage.

  • a) What is the probability of opening the garage?
  • b) If the chosen key allows to open the garage, find the probability that it belongs to the keyring A.

I answered

Let X be the "open garage" event:

  • $ P(X \cap A) = 1/5$
  • $ P(X \cap B) = 1/7$
  • $ P(X \cap C) = 1/8$

a) $P(X) = P(X \cap A) + P(X \cap B) + P(X \cap C) = \dfrac{131}{280}$

b) $P(A/X) = \dfrac{P(A \cap X)}{P(X)} = \dfrac{56}{131}$

Nowhere does it tell me that a key opens knowing it's from a keyring (conditional). But it also does not tell me that a key opens and that it is a certain keyring (intersection), however I chose this default option.

Then: How to know what to assume in cases like these?

Thank you very much.

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    Hint: if there had been a key ring D with one key, would $P(X\cap D) = 1$? – 2017-02-21
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    **Under the condition** that keyring A is chosen the probability the garage opens is $\frac15$. – 2017-02-21
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    In fact, you have $P(X\color{red}{\mid A})=\frac{1}{5}$, not $P(X\cap A)$. You have $P(X\cap A)=P(A)P(X\mid A) = \frac{1}{3}\cdot \frac{1}{5}$. Similarly, the others are also off by a factor of $\frac{1}{3}$. As for solving $P(A\mid X)$, use bayes' theorem to note $P(A\mid X)=\frac{P(A)P(X\mid A)}{P(X)}$ – 2017-02-21
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    @MeesdeVries In that case the probability would be greater than one. Then everything I did is wrong. How should I interpret the information given to me in the exercise? Thank you very much. – 2017-02-21

1 Answers 1

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First, note that for your first three probabilities, it's actually $\mathbb{P}[X|A]=\frac{1}{5}$, etc. Intersection means both events must be true, whereas when you calculated that you assumed you were "living in the universe" where $A$ is true and used that to find the probability of $X$ being true.

In general it's hard to answer how you should know which to use, it's a vague question. The only thing I can say is that it gets easier with practice.

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    It is correct to say that $ P (X) = P (X | A) * P (A) + P (X | B) * P (B) + P (X | C) * P (C) $? – 2017-02-21
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    Yes, that's exactly right. It comes from the equation $\mathbb{P}[X]=\mathbb{P}[X\cap A]+\mathbb{P}[X\cap B]+\mathbb{P}[X\cap C]$, because $A$, $B$, $C$ are known as a partition. – 2017-02-21